Question

Two discs have the same moment of inertia about their axis. Their thicknesses are \(t_1\) and \(t_2\) and they have the same density. If \( \dfrac{R_1}{R_2} = \dfrac{1}{2} \), find \( \dfrac{t_1}{t_2} \).

• A. \( \dfrac{1}{16} \)
• B. \( 16 \)
• C. \( \dfrac{1}{4} \)
• D. \( 4 \)

Hint:

For objects of the same material:
Always express mass in terms of density and volume
Look for proportional relationships to simplify calculations

Answer 1

The Correct Option is B

Concept:
The moment of inertia of a uniform solid disc about its central axis is: \[ I = \frac{1}{2}MR^2 \] For a disc of density \(\rho\), radius \(R\), and thickness \(t\): \[ M = \rho \times (\text{Volume}) = \rho \pi R^2 t \] Hence, \[ I = \frac{1}{2}\rho \pi R^4 t \] Thus, for discs of the same material: \[ I \propto R^4 t \]
Step 1: Use the given condition of equal moments of inertia. \[ I_1 = I_2 \Rightarrow R_1^4 t_1 = R_2^4 t_2 \]
Step 2: Substitute the given radius ratio. \[ \frac{R_1}{R_2} = \frac{1}{2} \Rightarrow \left(\frac{R_1}{R_2}\right)^4 = \frac{1}{16} \] \[ \frac{1}{16} \cdot t_1 = t_2 \]
Step 3: Find the required ratio. \[ \frac{t_1}{t_2} = 16 \] \[ \boxed{\dfrac{t_1}{t_2} = 16} \]